Assuming that petrol is octane ($C_8$$H_{18}$) and has density \[\text{ 0}\text{.8 g m}{{\text{L}}^{\text{-1}}}\] ,\[\text{ 1}\text{.425 litre}\] of petrol incomplete combustion will consume:
A).\[\text{ 50 mole of }{{\text{O}}_{\text{2}}}\]
B).\[\text{ 100 mole of }{{\text{O}}_{\text{2}}}\]
C).\[\text{ 125 mole of }{{\text{O}}_{\text{2}}}\]
D).\[\text{ 200 mole of }{{\text{O}}_{\text{2}}}\]

Hint: The hydrocarbons with the molecular formula \[{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}\] undergoes the complete combustion in presence of an excess of oxygen to produce the \[\text{C}{{\text{O}}_{\text{2}}}\] and \[{{\text{H}}_{\text{2}}}\text{O}\]. Here as the oxygen is supplied in the excess amount the number of moles of product formed will depend on the number of moles of propane undergoing the reaction.

Complete answer:
All hydrocarbons burn more than oxygen to form carbon dioxide and water molecules. The general reaction for the combustion of alkanes is as shown below:
\[{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}\text{ + }\left( \text{x+}\dfrac{\text{y}}{\text{2}} \right)\text{ }{{\text{O}}_{\text{2}}}\text{ }\to \text{ x C}{{\text{O}}_{\text{2}}}\text{ + }\dfrac{\text{y}}{\text{2}}\text{ }{{\text{H}}_{\text{2}}}\text{O}\]
In the general reaction, the x stands for the number of the carbon atom of alkane and y stands for the number of a hydrogen atom bonded to the carbon.
We are provided with the following data:
The propane $\text{ (}{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}}\text{) }$ undergoes complete combustion.
The density of propane is \[\text{ 0}\text{.8 g m}{{\text{L}}^{\text{-1}}}\]
The volume of petrol that undergoes combustion is\[\text{ 1}\text{.425 litre}\].
Let’s first write down the combustion reaction for the propane using the general reaction. we have,
\[\begin{align}
  & {{\text{C}}_{8}}{{\text{H}}_{18}}\text{ + }\left( \text{8+}\dfrac{18}{\text{2}} \right)\text{ }{{\text{O}}_{\text{2}}}\text{ }\to \text{ 8 C}{{\text{O}}_{\text{2}}}\text{ + }\dfrac{18}{\text{2}}\text{ }{{\text{H}}_{\text{2}}}\text{O} \\
 & \text{ Or} \\
 & \text{ }{{\text{C}}_{8}}{{\text{H}}_{18}}\text{ + }\left( \dfrac{25}{\text{2}} \right)\text{ }{{\text{O}}_{\text{2}}}\text{ }\to \text{ 8 C}{{\text{O}}_{\text{2}}}\text{ + 9 }{{\text{H}}_{\text{2}}}\text{O} \\
\end{align}\]
That is one mole of propane reacts with \[\left( \dfrac{25}{\text{2}} \right)\] the number of moles of oxygen.
The mass of propane can be calculated as:
From $\text{ }\!\!\rho\!\!\text{ = }\dfrac{\text{M}}{\text{V}}\text{ }$ we get, $\text{ M = }\!\!\rho\!\!\text{ }\!\!\times\!\!\text{ V }$
Now substitute the values, we have
$\begin{align}
  & \text{ M = }\!\!\rho\!\!\text{ }\!\!\times\!\!\text{ V } \\
 & \text{ = ( 0}\text{.8 g m}{{\text{L}}^{\text{-1}}}\text{ ) }\!\!\times\!\!\text{ ( 1}\text{.425 liter )} \\
 & \text{ = ( 0}\text{.8 g m}{{\text{L}}^{\text{-1}}}\text{ ) }\!\!\times\!\!\text{ ( 1425 mL )} \\
 & \text{ = ( 1140 g )} \\
\end{align}$
Therefore, the mass of propane is $\text{1140 g}$.
Let's calculate the number of moles of propane. We have,
$\begin{align}
  & \text{ No of moles of propane = }\dfrac{\text{mass of propane }}{\text{Molecular weight of Propane }}\text{ } \\
 & \text{ = }\dfrac{\text{1140 g }}{\text{114 g mo}{{\text{l}}^{\text{-1}}}} \\
 & \text{ = 10}\text{.00 mol } \\
\end{align}$
Since the molecular weight of propane is $\text{114 g mo}{{\text{l}}^{\text{-1}}}$.
Now we can say that, if one of the moles of propane reacts with the \[\left( \dfrac{25}{\text{2}} \right)\] moles of oxygen, then $\text{10}\text{.96 mole}$ the propane reacts with ‘X’ number of moles of oxygen.
This can be written as:
$\begin{matrix}
   \text{1 mole of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}} & \text{=} & \dfrac{\text{25}}{\text{2}}\text{moles of }{{\text{O}}_{\text{2}}} \\
   \text{10}\text{.00 moles of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}}\text{ } & \text{=} & \text{X moles of }{{\text{O}}_{\text{2}}} \\
\end{matrix}\text{ }$
On cross multiplying we get,
\[\begin{align}
  & \text{ } \\
 & \begin{matrix}
   \left( \text{X moles of }{{\text{O}}_{\text{2}}} \right)\text{ }\times \text{ }\left( \text{1 mole of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}} \right) & = & \left( \dfrac{\text{25}}{\text{2}}\text{moles of }{{\text{O}}_{\text{2}}} \right)\text{ }\times \text{ }\left( \text{10}\text{.00 moles of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}}\text{ } \right) \\
\end{matrix} \\
 & \\
 & \text{Or }\left( \text{X moles of }{{\text{O}}_{\text{2}}} \right)\text{ = }\dfrac{\left( \dfrac{\text{25}}{\text{2}}\text{moles of }{{\text{O}}_{\text{2}}} \right)\text{ }\times \text{ }\left( \text{10}\text{.00 moles of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}}\text{ } \right)}{\left( \text{1 mole of }{{\text{C}}_{\text{8}}}{{\text{H}}_{\text{18}}} \right)} \\
 & \\
 & \text{Or }\left( \text{X moles of }{{\text{O}}_{\text{2}}} \right)\text{ = }\left( \text{ 125 moles of }{{\text{O}}_{\text{2}}} \right) \\
\end{align}\]

Thus the number of moles of oxygen reacts when the \[\text{ 1}\text{.425 litre}\] of petrol undergoes the combustion is equal to 125 moles.

Hence, (C) is the correct option.

Note:
- If there is an insufficient supply of oxygen it results in the incomplete combustion of a hydrocarbon. The incomplete combustion generates carbon monoxide as the product instead of carbon dioxide.
- In the case of complete combustion of hydrocarbons, the hydrocarbons are limiting reagents. The number of moles of carbon dioxide produced depends on the number of moles of hydrocarbon undergoing the reaction.